Answer :

Given: (1 + i)3 – (1 – i)3 …(i)

We know that,


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(a – b)3 = a3 – 3a2b + 3ab2 – b3


By applying the formulas in eq. (i), we get


(1)3 + 3(1)2(i) + 3(1)(i)2 + (i)3 – [(1)3 – 3(1)2(i) + 3(1)(i)2 – (i)3]


= 1 + 3i + 3i2 + i3 – [1 – 3i + 3i2 – i3]


= 1 + 3i + 3i2 + i3 – 1 + 3i – 3i2 + i3


= 6i + 2i3


= 6i + 2i(i2)


= 6i + 2i(-1) [ i2 = -1]


= 6i – 2i


= 4i


= 0 + 4i



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